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\Title Lemniscate.
   
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\centerline {\includegraphics[width=2.5in] {Lemniscate.png}}

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\lf
The Lemniscate is a figure-eight curve with a simple {\bf mechanical construction}
attributed to Bernoulli: Choose two `focal' points $F_1,F_2$ at distance $L$,
then take three rods, one of length $L$, two of length $L/\sqrt{2}$. The short
ones can rotate around the focal points and they are connected by the long one
with joints which allow rotation. This machine has one degree of freedom and
the {\it midpoint} of the long rod traces out the Lemniscate when one of the short rods
is rotated. {\it Other drawing pens} can be chosen by setting $bb \ne 0$, see the 
default {\tt Morph}.
\lf
Mechanical constructions of curves often come with simple {\bf tangent constructions}.
We imagine that a plane is attached to the long rod. Then every point of this plane
traces out a curve when the rods move. At each moment the endpoints of the long rod
move orthogonal to the short rods (namely on circles around the focal points). This
says that the straight extensions of the short rods (green) intersect in the momentary
center of rotation. At this moment every point of the plane rotates around this
center so that the tangent of each point's orbit is orthogonal to its connection with 
the momentary center of rotation. (Compare the other mechanically constructed curves.)
\lf
The Lemniscate has the implicit equation:\hfil\break\vskip-10pt
$(x^2 + y^2)^2 = x^2 - y^2$. 

\noindent
Divide this by $r^2:= x^2 + y^2$
to get the polar form:\hfil\break\vskip-10pt
$ r^2 = \cos(\phi)^2 - \sin(\phi)^2$.

\noindent
Parametrizations are not unique, here is one:\hfil\break\vskip-10pt
$x(t) := \cos(t)/(1+\sin(t)^2)$
  
$y(t) :=\sin(t)\ \cos(t)/(1+\sin(t)^2)$. 

\noindent
The points $F_1,F_2:=\pm1/\sqrt{2}$ are called Focal points of the
Lemniscate because of the special property:\par
$|P-F_1|\cdot|P-F_2| = |F_1-F_2|^2/4$. 
\vskip2pt \noindent
If one takes the complex square-root of a circle that touches
the $y$-axis from the right at $0$ then one also obtains a Lemniscate.
In the Conformal Category, choose $z \to \sqrt{z}$, and then in the
Action Menu, select Choose Circle by Mouse, and create a circle that
is tangent to the $y$-axis at $0$.)
\vskip2pt \noindent
The inversion map: $ (x,y)\mapsto (x,y)/(x^2+y^2)$ often transforms
some interesting curve into another interesting curve. Indeed, the
Lemniscate, with the above parametrization, is transformed by inversion 
into the curve: \par  $x =1/\cos(t),\ y=\sin(t)/\cos(t)$. 

\noindent
Since $x^2-y^2=1$, 
this is a hyperbola, so we could also have obtained the Lemniscate 
from the standard hyperbola by inversion. 

\noindent
We note that not every figure-eight curve is a Lemniscate, another
figure-eight is obtained by the simpler parametrization:


$x(t) := \cos(t)$

$y(t) := \sin(t)\cdot \cos(t),$ 

\noindent
which has the implicit equation $y^2 = x^2(1-x^2)$.

\noindent H.K.

\bye
